Can we get a negative value for areas of surfaces of revolution?

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SUMMARY

The discussion centers on calculating the surface area of a solid of revolution generated by the curve defined by the equation x = (1/3)y^(3/2) - y^(1/2) when revolved around the y-axis for the interval 1 ≤ y ≤ 3. The correct formula for surface area is S = 2π ∫ (radius)(dS), where dS is the differential of arc length. The user encountered negative values in their calculations due to an oversight in integrating the function correctly. The final surface area, when calculated properly, yields a positive result, confirming that area cannot be negative.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of surfaces of revolution.
  • Knowledge of differential arc length, represented as dS = √(1 + (dx/dy)²) dy.
  • Ability to manipulate and integrate polynomial expressions.
NEXT STEPS
  • Review the process of calculating surface areas using the formula S = 2π ∫ (radius)(dS).
  • Study the integration of functions involving square roots and polynomial expressions.
  • Learn about the properties of definite integrals and their implications for physical quantities like area.
  • Practice problems involving surfaces of revolution to reinforce understanding of the concepts discussed.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and geometric applications, as well as educators teaching these concepts.

joseph9496
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the question is..

Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1≦y≦3; revolved about y-axis

so i use the general formula "S = Integral 2π (radius)(dS)"
and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..
i just keep getting negative value when i subsitute everything in the equation...

thanks.
 
Last edited:
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joseph9496 said:
the question is..

Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1 <= y <= 3; revolved about y-axis

so i use then general formula "S = 2π INTEGRAL (radius)(dS)"
and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..
i just keep getting negative value when i subsitute everything in the equation...

thanks.

Yes, the surface area is given by 2\pi \int x dS. Here dS is the differential of arc length, \sqrt{1+ (dx/dy)^2} dy
Since x= (1/3)y^(3/2)- y^(1/2), dx/dy= (1/2)y^{1/2}- (1/2)y^{-1/2} and (dx/dy)^2= (1/4)(y- 2+ y^{-1}). Then (dx/dy)^2+ 1= (1/4)(y+ 2+ y^{-1})= (1/4)(y^{1/2}+ y^{-1/2})^2 so dS= \sqrt{(1/4)(y+ 2+ y^{-1})}= (1/2)(y^{1/2}+ y^{-1/2})dy.

The surface area is
\pi\int_1^3 (y^{1/2}+ y^{-1/2}) dy
That should be easy to integrate and you definitely won't get a negative value!

To answer you original question, no "area" is never negative.
 
HallsofIvy said:
The surface area is
\pi\int_1^3 (y^{1/2}+ y^{-1/2}) dy
That should be easy to integrate and you definitely won't get a negative value!

Yes i got to that part too but what happen to the radius (x)?
cuz after i multiply the radius with the dS and subsitute value 3 and 1 into y-value i got
-16/9..
 
Last edited:
You are right- I forgot to multiply by x= (1/3)y^{3/2}- y^{1/2}. That would make the integral
\pi \int_1^3 ((1/3)y^2- (2/3)y- 1)dy
 
yea..and after i do anti-derivative i got

π [ (1/9)y^3 - (2/6)y^2 -y ]

which gives me a -16/9...

i just dun understand how come my answer is negative..

btw thanks for helping!
 

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